Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $p = \dfrac{9x - 63}{x^2 + x - 56} \div \dfrac{9x - 9}{x + 8} $
Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{9x - 63}{x^2 + x - 56} \times \dfrac{x + 8}{9x - 9} $ First factor the quadratic. $p = \dfrac{9x - 63}{(x + 8)(x - 7)} \times \dfrac{x + 8}{9x - 9} $ Then factor out any other terms. $p = \dfrac{9(x - 7)}{(x + 8)(x - 7)} \times \dfrac{x + 8}{9(x - 1)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 9(x - 7) \times (x + 8) } { (x + 8)(x - 7) \times 9(x - 1) } $ $p = \dfrac{ 9(x - 7)(x + 8)}{ 9(x + 8)(x - 7)(x - 1)} $ Notice that $(x - 7)$ and $(x + 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 9(x - 7)\cancel{(x + 8)}}{ 9\cancel{(x + 8)}(x - 7)(x - 1)} $ We are dividing by $x + 8$ , so $x + 8 \neq 0$ Therefore, $x \neq -8$ $p = \dfrac{ 9\cancel{(x - 7)}\cancel{(x + 8)}}{ 9\cancel{(x + 8)}\cancel{(x - 7)}(x - 1)} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $p = \dfrac{9}{9(x - 1)} $ $p = \dfrac{1}{x - 1} ; \space x \neq -8 ; \space x \neq 7 $